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Trig proof

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    This always seemed intuitive to me, but when I tried to prove it I got stuck:

    sin(x +pi/2) = cos(x)

    It is easy with the angle addition formula, but is there another way?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 25, 2007 #2


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    Homework Helper

    how else do u want to prove it?

    e^{i(\theta + \pi/2)} = \cos (\theta + \pi/2) +i \sin (\theta + \pi/2) \qquad\quad (1)

    e^{i(\theta + \pi/2)} = e^{i\theta} e^{i\pi/2} = i e^{i\theta}
    = -\sin (\theta) + i \cos (\theta) \quad (2)

    equating Re and Im part of (1) and (2) to get two relationships between sin and cos.
  4. Oct 25, 2007 #3
    I love the beauty of this mathematical proof:

    Using Taylor expansion about [itex]x=0[/itex]:

    [tex]\cos(x) = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + ...[/tex]

    [tex]\sin(x) = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 + ...[/tex]

    [tex]e^{ix} = 1 + ix - \frac{1}{2}x^2 - i\frac{1}{6}x^3 + ... + \frac{i^n}{n!}x^n[/tex]

    [tex]e^{ix} = \cos(x) + i\sin(x)[/tex]

    It still amazes me, absolutely incredible :P

  5. Oct 26, 2007 #4


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    There are many ways. The analytical approaches are very nice, but awfully sophisticated and high-powered.

    This trig relation just came up in the work I'm doing with students on torque. Here's a trigonometric proof (which I'll describe rather than scanning and uploading a drawing):

    Draw a right triangle and mark one of the non-right angles, theta. The other angle is of course complimentary, so it's (90º - theta); the sine of this angle will be the cosine of the other angle, which is the familiar "co-relation"

    [tex] sin(90º - \theta) = cos \theta[/tex].

    Now extend the side of the triangle adjacent to the complimentary angle outward away from the right angle. The angle between that ray and the hypotenuse is supplementary to the angle (90º - theta), so its measure is 180º - (90º - theta) = 90º + theta . But the sine of a supplementary angle is the same as the sine of the angle itself:

    [tex] sin(180º - \theta) = sin \theta[/tex] , so

    [tex] sin(90º + \theta) = sin(90º - \theta) = cos \theta[/tex]. Q.E.D.

    P.S. *heh* I just thought of a graphical way to prove it. The graph of sin x looks like the graph of cos x translated to the right by [tex]\frac{\pi}{2}[/tex]. So if you shift the graph of sin x to the left by the same amount, you have [tex] sin( x + \frac{\pi}{2}) = cos x [/tex].
    Last edited: Oct 26, 2007
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